Second Order Linear Differential Equation (Non-constant coefficients)

SherlockOhms
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Homework Statement


Find the general solution for the following differential equation:
y'' + x(y')^2 = 0.


Homework Equations


Integration, differentiation...


The Attempt at a Solution


Usually for these sort of DE you could use the substitution v(x) = y'(x) and this would simplify such an equation to a first order DE. The (y')^2 part is throwing me off as this would give the equation:
v(x)' + x(v(x))^2 = 0.
This would be grand if it weren't for the v(x)^2 term. Any ideas on how to get around this? Thanks.
 
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Are you worried about the v^2 term because it makes the eqn non-linear? If so, there are ways to solve such an equation.
 
Yeah. We've only covered linear DE's.
 
DAPOS said:
Yeah. We've only covered linear DE's.

Are you at all familiar with separation of variables?
 
Yeah. So I'll have v'(x) = (-x)(v(x))^2 and I can then separate the variables and solve. Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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